Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2022 AMC 10B Problems/Problem 24"

(→‎Solution 4 (Translation))
m (→‎Solution 3 (Slopes))
 
(6 intermediate revisions by 3 users not shown)
Line 15: Line 15:
  
 
Note that
 
Note that
−
<cmath>\begin{align*}
+
<cmath>\begin{alignat*}{8}
−
|f(800)-f(300)|&\leq 250, \\
+
|f(800)-f(300)| &\leq \frac12|800-300| &&= 250, \\
−
|f(800)-f(900)|&\leq 50, \\
+
|f(800)-f(900)| &\leq \frac12|800-900| &&= 50, \\
−
|f(400)-f(300)|&\leq 50, \\
+
|f(400)-f(300)| &\leq \frac12|400-300| &&= 50, \\
−
|f(400)-f(900)|&\leq 250. \\
+
|f(400)-f(900)| &\leq \frac12|400-900| &&= 250. \\
−
\end{align*}</cmath>
+
\end{alignat*}</cmath>
−
Let <math>a=f(300)=f(900).</math> Together, we conclude that  
+
Let <math>a=f(300)=f(900).</math> Together, it follows that  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
 
|f(800)-a|&\leq 50, \\
 
|f(800)-a|&\leq 50, \\
Line 106: Line 106:
 
~Viliciri (LaTeX edits)
 
~Viliciri (LaTeX edits)
  
−
==Solution 3 (Educated Guess)==
+
==Solution 3 (Slopes)==
  
 
Divide both sides by <math>|x - y|</math> to get <math>\frac{|f(x) - f(y)|}{|x - y|} \leq \frac{1}{2}</math>. This means that when we take any two points on <math>f</math>, the absolute value of the slope between the two points is at most <math>\frac{1}{2}</math>.
 
Divide both sides by <math>|x - y|</math> to get <math>\frac{|f(x) - f(y)|}{|x - y|} \leq \frac{1}{2}</math>. This means that when we take any two points on <math>f</math>, the absolute value of the slope between the two points is at most <math>\frac{1}{2}</math>.
  
−
Let <math>f(300) = f(900) = c</math>, and since we want to find the maximum value of <math>|f(800) - f(400)|</math>, we can take the most extreme case and draw a line with slope <math>\frac{-1}{2}</math> down from <math>f(300)</math> to <math>f(400)</math> and a line with slope <math>\frac{-1}{2}</math> up from <math>f(900)</math> to <math>f(800)</math>. Then <math>f(400) = c - 50</math> and <math>f(800) = c + 50</math>, so <math>|f(800) - f(400)| = |c + 50 - (c - 50)| = 100</math>, and this is attainable because the slope of the line connecting <math>f(400)</math> and <math>f(800)</math> still has absolute value less than <math>\frac{1}{2}</math>.
+
Let <math>f(300) = f(900) = c</math>, and since we want to find the maximum value of <math>|f(800) - f(400)|</math>, we can take the most extreme case and draw a line with slope <math>-\frac{1}{2}</math> down from <math>f(300)</math> to <math>f(400)</math> and a line with slope <math>\frac{1}{2}</math> up from <math>f(800)</math> to <math>f(900)</math>. Then <math>f(400) = c - 50</math> and <math>f(800) = c + 50</math>, so <math>|f(800) - f(400)| = |c + 50 - (c - 50)| = 100</math>, and this is attainable because the slope of the line connecting <math>f(400)</math> and <math>f(800)</math> still has absolute value less than <math>\frac{1}{2}</math>.
  
 
Therefore, <math>|f(f(800)) - f(f(400))| \leq \frac{1}{2}|f(800) - f(400)| = \frac{1}{2}(100) = \boxed{\textbf{(B)}\ 50}</math>.
 
Therefore, <math>|f(f(800)) - f(f(400))| \leq \frac{1}{2}|f(800) - f(400)| = \frac{1}{2}(100) = \boxed{\textbf{(B)}\ 50}</math>.
Line 116: Line 116:
 
==Solution 4 (Translation)==
 
==Solution 4 (Translation)==
  
−
Consider <math> g(x) = f(x)-f(300) </math>. Then <math> g(x)</math> satisfies all the conditions and <math> g(300) = g(900) = 0 </math>. We would want <math> g(400)</math> and <math> g(800)</math> as distant from each other as possible. So assign <math> g(400) = -50</math> and <math> g(800) = 50</math>, the possible lower and upper bounds respectively. It follows that one can obtain the upper bound for <math>|g(g(800)) - g(g(400))| = |g(50) - g(-50  )| \leq \frac{1}{2} \frac{1}{2}(100) = \boxed{\textbf{(B)}\ 50}</math> as the answer.
+
Consider <math> g(x) = f(x)-f(300) </math>. Then <math> g(x)</math> satisfies all the conditions and <math> g(300) = g(900) = 0 </math>. We would want <math> g(400)</math> and <math> g(800)</math> as distant from each other as possible. So assign <math> g(400) = -50</math> and <math> g(800) = 50</math>, the possible lower and upper bounds respectively. It follows that one can obtain the upper bound for <math>|g(g(800)) - g(g(400))| = |g(50) - g(-50  )| \leq \frac{1}{2}(100) = \boxed{\textbf{(B)}\ 50}</math> as the answer.
  
 
==Video Solution==
 
==Video Solution==
Line 128: Line 128:
  
 
~ pi_is_3.14
 
~ pi_is_3.14
 +
 +
==Video Solution by Interstigation==
 +
https://youtu.be/n0ZTR6edOBs
 +
 +
~Interstigation
  
 
==See Also==
 
==See Also==

Latest revision as of 18:53, 4 November 2023

Problem

Consider functions $f$ that satisfy \[|f(x)-f(y)|\leq \frac{1}{2}|x-y|\] for all real numbers $x$ and $y$. Of all such functions that also satisfy the equation $f(300) = f(900)$, what is the greatest possible value of \[f(f(800))-f(f(400))?\] $\textbf{(A)}\ 25 \qquad\textbf{(B)}\ 50 \qquad\textbf{(C)}\ 100 \qquad\textbf{(D)}\ 150 \qquad\textbf{(E)}\ 200$

Solution 1 (Absolute Values and Inequalities)

We have \begin{align*} |f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| &&(\bigstar) \\ &\leq \frac12\left|\frac12|800-400|\right| \\ &= 100, \end{align*} from which we eliminate answer choices $\textbf{(D)}$ and $\textbf{(E)}.$

Note that \begin{alignat*}{8} |f(800)-f(300)| &\leq \frac12|800-300| &&= 250, \\ |f(800)-f(900)| &\leq \frac12|800-900| &&= 50, \\ |f(400)-f(300)| &\leq \frac12|400-300| &&= 50, \\ |f(400)-f(900)| &\leq \frac12|400-900| &&= 250. \\ \end{alignat*} Let $a=f(300)=f(900).$ Together, it follows that \begin{align*} |f(800)-a|&\leq 50, \\ |f(400)-a|&\leq 50. \\ \end{align*} We rewrite $(\bigstar)$ as \begin{align*} |f(f(800))-f(f(400))| &\leq \frac12|f(800)-f(400)| \\ &= \frac12|(f(800)-a)-(f(400)-a)| \\ &\leq \frac12|50-(-50)| \\ &=\boxed{\textbf{(B)}\ 50}. \end{align*} ~MRENTHUSIASM

Solution 2 (Lipschitz Condition)

Denote $f(900)-f(600) = a$. Because $f(300) = f(900)$, $f(300) - f(600) = a$.

Following from the Lipschitz condition given in this problem, $|a| \leq 150$ and \[ f(800) - f(600) \leq \min \left\{ a + 50 , 100 \right\} \] and \[ f(400) - f(600) \geq \max \left\{ a - 50 , -100 \right\} . \] Thus, \begin{align*} f(800) - f(400) & \leq \min \left\{ a + 50 , 100 \right\} - \max \left\{ a - 50 , -100 \right\} \\ & = 100 + \min \left\{ a, 50 \right\} - \max \left\{ a , - 50 \right\} \\ & = 100 + \left\{ \begin{array}{ll} a + 50 & \mbox{ if } a \leq -50 \\ 0 & \mbox{ if } -50 < a < 50 \\ -a + 50 & \mbox{ if } a \geq 50 \end{array} \right. . \end{align*} Thus, $f(800) - f(400)$ is maximized at $a = 0$, $f(800)-f(600) = 50$, $f(400)-f(600)=-50$, with the maximal value 100.

By symmetry, following from an analogous argument, we can show that $f(800) - f(400)$ is minimized at $a = 0$, $f(800)-f(600) = -50$, $f(400)-f(600)=50$, with the minimal value $-100$.

Following from the Lipschitz condition, \begin{align*} f(f(800)) - f(f(400)) & \leq \frac{1}{2} \left| f(800) - f(400) \right| \\ & \leq 50 . \end{align*} We have already construct instances in which the second inequality above is augmented to an equality.

Now, we construct an instance in which the first inequality above is augmented to an equality.

Consider the following piecewise-linear function: \[ f(x) = \left\{ \begin{array}{ll} \frac{1}{2} \left( x - 300 \right) & \mbox{ if } x \leq 300 \\ -\frac{1}{2} \left( x - 300 \right) & \mbox{ if } 300 < x \leq 400 \\ \frac{1}{2} \left( x - 600 \right) & \mbox{ if } 400 < x \leq 800 \\ -\frac{1}{2} \left( x - 900 \right) & \mbox{ if } x > 800 \end{array} \right.. \] Therefore, the maximum value of $f(f(800)) - f(f(400))$ is $\boxed{\textbf{(B)}\ 50}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

~Viliciri (LaTeX edits)

Solution 3 (Slopes)

Divide both sides by $|x - y|$ to get $\frac{|f(x) - f(y)|}{|x - y|} \leq \frac{1}{2}$. This means that when we take any two points on $f$, the absolute value of the slope between the two points is at most $\frac{1}{2}$.

Let $f(300) = f(900) = c$, and since we want to find the maximum value of $|f(800) - f(400)|$, we can take the most extreme case and draw a line with slope $-\frac{1}{2}$ down from $f(300)$ to $f(400)$ and a line with slope $\frac{1}{2}$ up from $f(800)$ to $f(900)$. Then $f(400) = c - 50$ and $f(800) = c + 50$, so $|f(800) - f(400)| = |c + 50 - (c - 50)| = 100$, and this is attainable because the slope of the line connecting $f(400)$ and $f(800)$ still has absolute value less than $\frac{1}{2}$.

Therefore, $|f(f(800)) - f(f(400))| \leq \frac{1}{2}|f(800) - f(400)| = \frac{1}{2}(100) = \boxed{\textbf{(B)}\ 50}$.

Solution 4 (Translation)

Consider $g(x) = f(x)-f(300)$. Then $g(x)$ satisfies all the conditions and $g(300) = g(900) = 0$. We would want $g(400)$ and $g(800)$ as distant from each other as possible. So assign $g(400) = -50$ and $g(800) = 50$, the possible lower and upper bounds respectively. It follows that one can obtain the upper bound for $|g(g(800)) - g(g(400))| = |g(50) - g(-50 )| \leq \frac{1}{2}(100) = \boxed{\textbf{(B)}\ 50}$ as the answer.

Video Solution

https://youtu.be/2Li0IYOQCFQ

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution by OmegaLearn Using Algebraic Manipulation

https://youtu.be/-yzpw6b3_KA

~ pi_is_3.14

Video Solution by Interstigation

https://youtu.be/n0ZTR6edOBs

~Interstigation

See Also

2022 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2022_AMC_10B_Problems/Problem_24&oldid=200766"